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OCR A Jun 2012 Paper 5 Q4
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![4 Nitrogen dioxide reacts with ozone as shown below. 2NO2(g) + O3(g) N2O5(g) + O2(g) (a) The kinetics of the reaction between NO2 and O3 was investigated and the following experimental results were obtained. experiment [NO2(g)] / mol dm3 [O3(g)] / mol dm3 initial rate / mol dm3 s1 0.00150 0.00225 0.00225 0.00250 0.00250 0.00500 4.80108 7.20108 1.44107 (i) Determine the rate equation and rate constant for the reaction of NO2(g) and O3(g). In your answer you should make clear how your conclusions fit with the experimental results.[8]OCR 2012<br />
(ii) Suggest a possible two-step mechanism for this reaction.[2] (b) The feasibility of the reaction between NO2 and O3 is influenced by the enthalpy change and entropy change of the reaction and the temperature. 2NO2(g) + O3(g) N2O5(g) + O2(g) H = 198 kJ mol1 S = 168 J K1 mol1 (i) Explain why this reaction has a negative entropy change.[2] (ii) Calculate the value of G, in kJ mol1, at 25 C for the reaction of NO2 with O3. (iii) State and explain how the feasibility of this reaction will change with increasing temperature. G =kJ mol1 [3][2]OCR 2012 [Total: 17] Turn over<br />](https://learnah.org/wp-content/uploads/2018/08/OCR_5_QP_JUN12_question_4-0.jpg)
![F325 Mark Scheme June 2012 Question (a) (i) Answer initial rates data: From Experiment 1 to Experiment 2 AND [NO2] x 1.5, rate x 1.5 (cid:57) 1st order with respect to NO2 (cid:57) From Experiment 2 to Experiment 3 AND [O3] is doubled, rate x 2 (cid:57) 1st order with respect to O3 (cid:57) rate equation and rate constant: rate = k[NO2] [O3] (cid:57) OR k = rate [NO2][O3] 4.80 (cid:117)10(cid:16)8 0.00150 (cid:117) 0.00250 = 0.0128 (cid:57) dm3 mol1 s1 (cid:57) Marks Guidance ANNOTATIONS MUST BE USED Quality of Written Communication: -------------------------------------------------------------------------------- Changes MUST be linked to Experiment numbers in writing (Could be described unambiguously) IGNORE annotations in the table -------------------------------------------------------------------------------- For 2nd condition, ALLOW when [NO2] increases by half, rate increases by half NOTE: Orders may be identified within a rate equation ALLOW: working from any of the Experiments : All give the same calculated answer 0.0128 subsumes previous rearrangement mark ALLOW: mol1 dm3 s1 (cid:57) DO NOT ALLOW 0.013 ----------------------------------------------------------------------------- ALLOW ECF from inverted k expression: k = : k = 78.125 (cid:57) ALLOW 3 SF or more NOTE units must be from rate equation (cid:57) 2][O3] rate over-rounding [NO (cid:57) 14<br />
F325 Mark Scheme June 2012 Question Answer (a) (ii) step 1: NO2 + O3 LHS of step one (cid:57) (cid:127)(cid:111) NO3 + O2 step 2: NO2 + NO3 (cid:127)(cid:111) N2O5 rest of equations for step 1 AND step 2 (cid:57) CHECK that each equation is balanced CARE: Step 1 AND Step 2 must add up to give overall equation In Step 2, IGNORE extra species shown on both sides, e.g. NO2 + NO3 + O2 (cid:127)(cid:111) N2 Step 2 can only gain a mark when Step 1 is correct 3 gaseous moles (cid:127)(cid:111) 2 gaseous moles (cid:57) Less randomness OR becomes more ordered (cid:57) 5 + O2 (i) (ii) FIRST, CHECK THE ANSWER ON ANSWER LINE IF answer = 148 award 3 marks -------------------------------------------------------------------- G = HTS (cid:57) = 198(298 x 168/1000) (cid:57) = 148 (kJ mol1) (cid:57) (b) Marks Guidance (cid:127)(cid:111) NO5 State symbols NOT required For rest of equations, ALLOW other combinations that together give the overall equation, e.g.: NO2 + NO5 (cid:127)(cid:111) N2O5 + O2 e.g.: NO + NO2 + O2 (cid:127)(cid:111) N2O5 DO NOT ALLOW use of algebraic species, e.g. X (cid:127)(cid:111) NO + 2O2 ALLOW products have fewer gaseous moles ORA ALLOW molecules instead of moles ALLOW fewer ways of distributing energy OR fewer degrees of freedom OR fewer ways to arrange IF there is an alternative answer, check calculator value and working for intermediate marks by ECF ------------------------------------------------------------ 2nd mark subsumes 1st mark for G = HTS ALLOW 148 to calculator value of 147.936 ALLOW for 2 marks: 49866 (kJ mol1): not converting S from J to kJ (no1000) 193.8 (kJ mol1) use of 25 instead of 298 15<br />
F325 Mark Scheme June 2012 Question Answer (b) (iii) CARE: responses involve changes of negative values -------------------------------------------------------------------------- Feasibility with increasing temperature Reaction becomes less feasible/not feasible AND G increases OR G becomes less negative OR G = 0 OR G > 0 OR G is positive OR G approaches zero (cid:51) ***IF a candidate makes a correct statement about the link between G and feasibility, IGNORE an incorrect H and TS relationship IF there is no G statement, then mark any H and TS relationship in line with the mark scheme --------------------- Effect on TS TS becomes more negative OR TS decreases OR TS increases OR magnitude of TS increases (cid:51) --------------------------------------------------------------------------------- Total 17 16 Marks Guidance ANNOTATIONS MUST BE USED -------------------------------------------------------------------------------- As alternative for not feasible ALLOW not spontaneous OR a comment that implies reaction does not take place ALLOW for G increases H = TS OR H > TS OR HTS is positive OR TS becomes more significant than H OR TS becomes the same as H OR TS becomes more negative than H NOTE Last statement will also score 2nd mark ------------------- DO NOT ALLOW TS increases ---------------------------------------------------------------------- APPROACH BASED ON TOTAL ENTROPY: Feasibility with increasing temperature Reaction becomes less feasible/not feasible AND SH/T OR Stotal decreases/ less positive OR S outweighs/ is less significant than H/T (cid:51) Effect on H/T H/T is less negative OR H/T increases OR H/T decreases OR magnitude of H/T decreases (cid:51)<br />](https://learnah.org/wp-content/uploads/2018/08/OCR_5_QP_JUN12_answer_4-0.jpg)
